Question: $\begin{aligned} &H(x)=-x-5 \\\\ &h(x)=H'(x) \end{aligned}$ $\int_{-3}^{6} h(x)\,dx=$
Answer: $h$ is the derivative of $H$, which means $H$ is an antiderivative of $h$. Since we know the antiderivative of $h$, we can use the fundamental theorem of calculus: For every function $h$ and its antiderivative $H$, $\int_a^b h(x)\,dx=H(b)-H(a)$. $\begin{aligned} &\phantom{=}\int_{-3}^{6} h(x)\,dx \\\\ &=H({6})-H({-3}) \\\\ &=[-{6}-5]-[-{(-3)}-5] \\\\ &=-11-(-2) \\\\ &=-9 \end{aligned}$ In conclusion, $\int_{-3}^{6} h(x)\,dx=-9$